Calculate the maximum value of the angle of incidence that a ray can make with the axis of a fiber such that it is guided through the fiber for the following fiber parameters. $$ \text { (i) } n_1=1.6, n_2=1.5 \text {, (ii) } n_1=2.1, n_2=1.5 $$

Solution:

We know that for total internal reflection of a ray the angle of incident is given by,

$ \sin \alpha=\sqrt{n_1^2-n_2^2} \text { or } \alpha=\sin ^{-1}\left(n_1^2-n_2^2\right)^{1 / 2} $,

(i) $n_1=1.6, n_2=1.5$

$ \alpha=\sin ^{-1} \sqrt{2.56-2.25} $

$ \alpha \neq 33.83^{\circ} $

(ii) $n_1=2.1, n_2=1.5$

$ \alpha=\sin ^{-1} \sqrt{4.41-2.25} $

$ \dot{\alpha}=\infty $

$\dot{\alpha}=\infty$

This means that total internal reflection is possible at all the angles, i.e., $0^{\circ}\lt$ $\alpha\lt90^{\circ}$

The angle $\alpha$ defines the light-gathering power of a fiber and is measured in terms of the numerical aperture (NA) of the fiber, i.e.,

$ \mathrm{NA}=\sqrt{n_1^2-n_2^2} $

In first case, $\mathrm{NA}=0.55$ while is second case, $\mathrm{NA}=1.46$